How to solve this ???

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1 Answer
Jun 3, 2017

We have a=1 and b=2. For details please see below.

Explanation:

We have 3x^2+6x+5

= 3(x^2+2x)+5

= 3(x^2+2xx x xx1+1^2)-3xx1^2+5

= 3(x+1)^2-3+5

= 3(x+1)^2+2

Hence a=1 and b=2

As (x+1)^2 is a perfect square, and minimum value of 3x^2+6x+5 is 2 and hence it is always non-zero and positive and hence 2/(3x^2+6x+5) >=0

3(x+1)^2>=0 or 3(x+1)^2+2>=2

i.e. 3x^2+6x+5 >=2 and dividing by 3x^2+6x+5, we get

0 <=2/(3x^2+6x+5) <= 1
graph{2/(3x^2+6x+5) [-10, 10, -5, 5]}